Question

A 4.80-kg block is at rest on a horizontal floor. If you push horizontally on the 4.80-kg block with a force of 12.0 N, it just starts to move.A 7.00-kg block is stacked on top of the 4.80-kg block. What is the magnitude F of the force, acting horizontally on the 4.80-kg block as before, that is required to make the two blocks start to move together?

Answer 1

See below

Step-by-step explanation:

static friction coef = 12/ ( 4.80 * 9.81 N ) = .25484

Now the mass is changed to 7 + 4.8 = 11.8 kg

.25484 = x / (11.8 * 9.81)

x = 29.5 N

Answer 2

Given

Initial situation

m1: mass one

m1 = 4.8 kg

F: horizontal force

F = 12 N

Second situation

m2: mass two

m2 = 7kg

Procedure

Let's start with the free body diagram

Summation on y-axis

`\begin{gathered} N-mg=0 \\ N=mg \end{gathered}`

Summation on x-axis

`\begin{gathered} F-F_f=0 \\ F=F_f \\ \end{gathered}`

We can see that the Friction is equal to the horizontal Force. Now, we can calculate the coefficient of friction

Second part:

We can now calculate the force required to move the two blocks together

`\begin{gathered} F=\mu_smg \\ F=0.255\cdot(4.8+7)\cdot9.8 \\ F=29.50\text{ N} \end{gathered}`